Question

$21.4 \,g\, XCl$ salt is dissolved in water and formed $2$ litres of solution. The $pH$ of the resultant solution was found to be $5 .$ Calculate the density (in $g / cc$ ) of solid salt $XCl$, if $XCl$ forms $NaCl$ type crystal having $R _{ X ^{+}}$(radius of $X ^{+}$) $=0.8 \,\mathring{A}$ and $R _{ Cl ^{-}}$(radius of $Cl ^{-}$) $=1.7 \,\mathring{A}$.
Given: $K _{ b }( XOH )=2 \times 10^{-5} ; N _{ A }=6 \times 10^{23}$ Use $10^{-0.7}=$ $0.20, \log 2=0.3$

Answer 1

$K _{ b \times OH }=2 \times 10^{-5}$
$XCl \Rightarrow$ Salt of weak base and strong acid.
$pH =-\frac{1}{2}\left[\log K _{ w }+\log C -\log K _{ b }\right]$
$6 =-\frac{1}{2}\left[\log 10^{-14}+\log C -\log \left(2 \times 10^{-5}\right)\right] $
$=-\frac{1}{2}\left[-14+\log C -\log 2-\log 10^{-5}\right] $
$=-\frac{1}{2}[-14+\log C -0.3+5] $
$ 5 =-\frac{1}{2}[-14.3+5+\log C ]$
$10=14.3-5-\log C$
$\log C=14.3-5-10$
$\log C=-0.7$
$C=10^{-0.7}=0.2$
$C=0.2$
$\frac{21.4 / M}{2}=0.2$
$M=\frac{21.4}{2 \times 0.2}=53.5$
$ a =2 r^{+}+2 r^{-} $
$=2\left(r^{+}+r^{-}\right)=2 \times(0.8+1.7) $
$=2 \times 2.5$
$ a =5 \,\mathring{A} $
$ \rho =\frac{n \times \text { mol } \text { wt }}{N_{A} \times a^{3}} $
$\rho =\frac{4 \times 53.5}{6 \times 10^{23} \times\left(5 \times 10^{-8}\right)^{3}} $
$=\frac{4 \times 53.5}{6 \times 10^{23} \times 125 \times 10^{-24}} $
$=2.853$
$=2.85 \,g / cm ^{3} $